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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Therapsid on February 02, 2019, 08:51:08 AM

Hi,
Does (the amount of energy released) per (unit charge transferred as electrons) in a general redox reaction have a special name, or is it simply called the redox potential of that reaction?
I think that this quantity (I'll just call it redox voltage for now) is a very important property of redox reactions in general, not just ones happening in solution, because it tells us how "strong" the redox reaction is. Is it true that the redox voltage also allows us to predict whether a given redox reaction will take place? (The advantage of the redox voltage over the enthalpy change of reaction would be that it doesn't depend on how much of the reaction takes place, so the part reactions don't have to be balanced with each other for the prediction to be made.)
Example:
Mg (s) + 1/2 O_{2} (g) :rarrow: MgO (s)
ΔH = 601.6 kJ mol^{1} (4 s.f.)
redox voltage = (ΔH/N_{A}) / (2e) = ((601.6 kJ mol^{1}) / (6.022×10^{23} mol^{1})) / (2 × 1.602×10^{19} C) = 3.118 V (4 s.f.)
2 Na (s) + 1/2 O_{2} (g) :rarrow: Na_{2}O (s)
ΔH = 418.0 kJ mol^{1} (4 s.f.)
redox voltage = (ΔH/N_{A}) / (2e) = ((418.0 kJ mol^{1}) / (6.022×10^{23} mol^{1})) / (2 × 1.602×10^{19} C) = 2.166 V (4 s.f.)
3.118 > 2.166, therefore the reaction
Mg (s) + Na_{2}O (s) :rarrow: MgO (s) + 2 Na (s)
is exothermic (a little surprising, I know).
In addition, the redox voltage of this third reaction = redox voltage of 1^{st} reaction  redox voltage of 2^{nd} reaction = 3.118 V  2.166 V = 0.952 V (3 d.p.).
If any of my assumptions, calculations and/or conclusions are incorrect, please tell me.
Thanks in advance,
Therapsid

The redox potential is not related to the enthalpy change, but to the Gibbs energy change (which takes into account enthalpy and entropy, by the relation ΔG = ΔH  TΔS).
E is related to ΔG by the equation ΔG = nFE.
Therefore your calculated example is invalid.
ΔG°_{f} for many compounds is tabulated. (For elements in standard states it is zero.) I find values of 569.4 kJ/mol for MgO and 379.1 kJ/mol for Na_{2}O. Try your calculation again.

Thanks for your answer!
Applying your correction ΔG = nFE = nN_{A}eE to my calculation:
Mg (s) + 1/2 O_{2} (g) :rarrow: MgO (s) (1)
ΔG^{⦵}_{r1} = 569.4 kJ mol^{1} (4 s.f.)
n_{1} = 2
E^{⦵}_{r1} = ΔG^{⦵}_{r1} / (n_{1} N_{A} e) = (569.4 kJ/mol) / (2 × 6.022...×10^{23} mol^{1} × 1.602...×10^{19} C) = 2.950707... V = 2.951 V (4 s.f.)
2 Na (s) + 1/2 O_{2} (g) :rarrow: Na_{2}O (s) (2)
ΔG^{⦵}_{r2} = 379.1 kJ mol^{1} (4 s.f.)
n_{2} = 2
E^{⦵}_{r2} = ΔG^{⦵}_{r2} / (n_{2} N_{A} e) = (379.1 kJ/mol) / (2 × 6.022...×10^{23} mol^{1} × 1.602...×10^{19} C) = 1.964547... V = 1.965 V (4 s.f.)
Mg (s) + Na2O (s) :rarrow: MgO (s) + 2 Na (s) (3)
ΔG^{⦵}_{r3} = ΔG^{⦵}_{r1} − ΔG^{⦵}_{r2} = (569.4 kJ mol^{1}) − (379.1 kJ mol^{1}) = 190.3 kJ mol^{1} (4 s.f.)
n_{3} = n_{1} = n_{2} = 2 =: n
E^{⦵}_{r3} = ΔG^{⦵}_{r3} / (n_{3} N_{A} e) = (190.3 kJ/mol) / (2 × 6.022...×10^{23} mol^{1} × 1.602...×10^{19} C) = 0.986160... V
= (ΔG^{⦵}_{r1} − ΔG^{⦵}_{r2}) / (n N_{A} e) = (ΔG^{⦵}_{r1} / (n N_{A} e)) − (ΔG^{⦵}_{r2} / (n N_{A} e)) = E^{⦵}_{r1} − E^{⦵}_{r2} = 2.950707... V − 1.964547... V = 0.986160... V
= 0.986 V (3 d.p.)
We see that, as expected, the reaction is indeed thermodynamically favorable.

That is correct; the only quibble I have with your presentation is writing e.g. "569.4 kJ/mol" in your calculations, rather than "569,400 J/mol". If you multiply out the numbers you have written, the answer will be wrong. Obviously you have remembered to make the conversion, but to avoid mistakes it is helpful to write it out explicitly, and also in an exam will make your working clearer to the examiner.

Thanks!